Question

A charged particle of radius $5\times {10}^{-7}$ m is loacted in a horizontal electric field of intensity $6.28\times {10}^{5}V{m}^{-1}$. The surrounding medium has the coefficient of viscosity $\eta =1.6\times {10}^{5}Ns{m}^{-2}$. The particle starts moving under the effect of electric field and finally attains a uniform horizontal speed of $0.02m{s}^{-1}$. Find the number of electrons on it :

(Assume gravity free space)

• A. $3\times {10}^{11}$
• B. $6\times {10}^{11}$
• C. $9\times {10}^{11}$
• D. $1\times {10}^{11}$

Answer 1

The correct option is A $3\times {10}^{11}$

Using Stoke's law, $F=6\pi \eta rv$
here, $F=qE=NeE=6\pi \eta rv$
or $N=\frac{6\pi \eta rv}{eE}=\frac{6\times 3.14\times (1.6\times {10}^5)\times (5\times {10}^{-7})\times 0.02}{1.6\times {10}^{-19}\times 6.28\times {10}^5}=3\times {10}^{11}$